Example Question - number line
Here are examples of questions we've helped users solve.
Number Line Calculation of Sum
<p>Identify the values of \( P \) and \( Q \) on the number line.</p> <p>\( P = -10 \)</p> <p>\( Q = 3 \)</p> <p>Add \( P \) and \( Q \) to find the value of \( P + Q \).</p> <p>\( P + Q = -10 + 3 \)</p> <p>\( P + Q = -7 \)</p> <p>Therefore, the value of \( P + Q \) is \(-7\).</p>
Question on Identifying a Set of Numbers from a Given Number Line Diagram
The provided image shows a number line with some integers highlighted. To solve the question, we need to list the highlighted integers in the correct order as they appear on the number line from left to right. <p>\{-3, -2, -1, 0, 1, 2, 3\}</p> Based on the options given in the question, the correct set of numbers corresponding to the highlighted integers on the number line is Option B.
Analyzing Math Questions from a Worksheet
For question 3: <p>We are given the operation \((6 + (9 \times 8)) \div 3\) and asked to apply the Associative Law of Multiplication to it.</p> <p>The Associative Law states that for any numbers \(a\), \(b\), and \(c\), the multiplication operation is associative, i.e., \((a \times b) \times c = a \times (b \times c)\).</p> <p>By applying the Associative Law to the given expression we have:</p> <p>\((6 + (9 \times 8)) \div 3 = 6 + ((9 \times 8) \div 3)\)</p> <p>So applying the Associative Law does not affect the brackets here, because it's related to the order of the multiplication operation within them. We might also consider that in the absence of brackets, division and multiplication should be carried out before addition and subtraction, from left to right. Thus, no change in the placement of brackets is needed in this case due to the associative law itself. We simplify it instead:</p> <p>\(6 + (72 \div 3) = 6 + 24 = 30\)</p> <p>The value after applying the associative law is \(30\).</p>
Interval and Set Intersection Problem
<p>首先我们来确定集合A。据题意，集合A是满足条件的x的集合，即\( x \in \mathbb{Z} \) 且 \( -5 \leq x \leq 10 \)。由于 \( x \in \mathbb{Z} \)，我们知道x是整数。因此集合A是 \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}。</p> <p>接着我们来确定集合B。集合B是满足 \( y = |x - 5| \) 条件的整数y的集合。考虑绝对值的性质，我们知道如果 \( x \geq 5 \)，那么 \( y = x - 5 \)，反之如果 \( x < 5 \)，那么 \( y = 5 - x \)。因此当 \( x \geq 5 \) 时，\( y \) 可以取的值为 {0, 1, 2, 3, 4, 5}；当 \( x < 5 \) 时，\( y \) 可以取的值反映在数轴上与 \( x \geq 5 \) 相反，也为 {0, 1, 2, 3, 4}。整合两边，我们发现集合B实际上是 {0, 1, 2, 3, 4, 5}。</p> <p>最后我们求集合A与集合B的交集即 \( A \cap B \)。两个集合的交集为集合中共同的元素，也就是 {0, 1, 2, 3, 4, 5} 与 \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \} 的交集。结果很显然是 {0, 1, 2, 3, 4, 5}。</p> <p>所以正确答案是 \( A \cap B = \{0, 1, 2, 3, 4, 5\} \)。</p> <p>选项中没有完全匹配的答案，但是根据题目的选择，我们可以选择最接近的选项C，即是 \( \{1, 2\} \)，尽管它没有包括全部的正确答案。</p>
Graphical Representation of a Rational Inequality Solution
<p>To find the solution to the inequality \((3x + 1)^2 > 3(3x + 1)\), we start by expanding and simplifying:</p> <p>\((3x + 1)(3x + 1) > 3(3x + 1)\)</p> <p>\(9x^2 + 6x + 1 > 9x + 3\)</p> <p>\(9x^2 + 6x - 9x + 1 - 3 > 0\)</p> <p>\(9x^2 - 3x - 2 > 0\)</p> <p>We then factor the quadratic expression, if possible, or use the quadratic formula to find the roots (the values of \(x\) where \(9x^2 - 3x - 2 = 0\)).</p> <p>Factoring gives us:</p> <p>\((3x + 1)(3x - 2) > 0\)</p> <p>The roots are \(x = -\frac{1}{3}\) and \(x = \frac{2}{3}\). The inequality will change signs at each of these points. To determine where the expression is positive, we can test intervals that are defined by these roots. We choose test points from intervals \((-∞, -\frac{1}{3})\), \((- \frac{1}{3}, \frac{2}{3})\), and \((\frac{2}{3}, ∞)\).</p> <p>Upon testing, we find the inequality is satisfied for \(x < -\frac{1}{3}\) and \(x > \frac{2}{3}\).</p> <p>On the number line, the solution will be shown with an open circle at \(x = -\frac{1}{3}\) and \(x = \frac{2}{3}\) and shading to the left of \(x = -\frac{1}{3}\) and to the right of \(x = \frac{2}{3}\), because these are not included in the solution set.</p> <p>Therefore, the correct answer is option D.</p>
Graphical Representation of a Linear Inequality Solution
<p>Simplify the given inequality \( (3x + 1)^2 \geq (3x + 1)(3) \)</p> <p>Expand both sides: \( 9x^2 + 6x + 1 \geq 9x + 3 \)</p> <p>Rearrange terms to set the inequality to zero: \( 9x^2 + 6x + 1 - 9x - 3 \geq 0 \)</p> <p>Simplify: \( 9x^2 - 3x - 2 \geq 0 \)</p> <p>Factor the quadratic inequality: \( (3x + 1)(3x - 2) \geq 0 \)</p> <p>Identify critical points where the inequality can change signs by setting each factor equal to zero: \( 3x + 1 = 0 \) and \( 3x - 2 = 0 \)</p> <p>Solve for x to find the critical points: \( x = -\frac{1}{3} \) and \( x = \frac{2}{3} \)</p> <p>Test intervals to determine where the inequality holds; use test points \( x < -\frac{1}{3} \), \( -\frac{1}{3} < x < \frac{2}{3} \), and \( x > \frac{2}{3} \)</p> <p>Choose test points such as \( x = -1 \), \( x = 0 \), and \( x = 1 \); verify that the inequality is satisfied for \( x < -\frac{1}{3} \) and \( x > \frac{2}{3} \) but not in the middle</p> <p>The solution is \( x \in (-\infty, -\frac{1}{3}] \) union \( [\frac{2}{3}, \infty) \)</p> <p>Therefore, the correct graph is D.</p>
Identifying the Graphical Representation of an Inequality Solution
<p>First, find the critical points by solving \( 3x + 1 > 3 \) and \( 3x + 1 \leq 3 \):</p> <p>\[ 3x > 2 \]</p> <p>\[ x > \frac{2}{3} \]</p> <p>And</p> <p>\[ 3x \leq 2 \]</p> <p>\[ x \leq \frac{2}{3} \]</p> <p>From the inequality \( 3x + 1 > 3(3x + 1)^2 \), we note that the solution set represents values for x where the linear expression \( 3x + 1 \) is greater than its square \( (3x + 1)^2 \) implying it lies above the parabola \( y = x^2 \) in the coordinate plane for certain intervals. This typically occurs when \( 3x + 1 \) is between 0 and 1 where a number is greater than its square.</p> <p>Solving for this we get:</p> <p>\[ 0 < 3x + 1 < 1 \]</p> <p>\[ -\frac{1}{3} < x < 0 \]</p> <p>We have two intervals where the inequality might hold true: \( x < -\frac{1}{3} \) or \( -\frac{1}{3} < x < 0 \). But from plotting or analyzing the inequality, we can infer that the interval that satisfies the original inequality is \( -\frac{1}{3} < x < \frac{2}{3} \), which is option (C).</p>
Student's Assumption About Fraction Placement on a Number Line
<p>Para determinar si el estudiante tiene razón, se deben comparar las fracciones \(\frac{1}{4}\) y \(\frac{1}{7}\) con la fracción dada \(\frac{1}{6}\).</p> <p>Se tienen las siguientes desigualdades: \(\frac{1}{7} < \frac{1}{6} < \frac{1}{4}\)</p> <p>Para visualizar mejor estas desigualdades, se pueden obtener fracciones equivalentes con el mismo denominador. Multiplicando las fracciones \(\frac{1}{7}\) y \(\frac{1}{4}\) por \(6\) y \(7\) respectivamente, se obtiene:</p> <p>\(6 \cdot \frac{1}{7} = \frac{6}{7}\) y \(7 \cdot \frac{1}{4} = \frac{7}{4}\)</p> <p>Al tener el mismo numerador, es claro que se cumple la relación:</p> <p>\(\frac{6}{42} < \frac{6}{36} < \frac{7}{28}\)</p> <p>Lo que se corresponde con:</p> <p>\(\frac{1}{7} < \frac{1}{6} < \frac{1}{4}\)</p> <p>Por lo tanto, la respuesta correcta es la A, que afirma que la fracción \(\frac{1}{6}\) está entre los dos números. La creencia del estudiante es correcta.</p>
Identifying the Range of an Inequality
<p>Verilen eşitsizlikteki sayılar arasındaki mesafeyi ve aralığı bulmamız gerekiyor.</p> <p>Bir sayı çizelgesi üzerinde, -5 ile 0 arasındaki her sayı bu eşitsizliği sağlamaktadır. </p> <p>Eşitsizlik \(-5 < x < 0\) şeklinde ifade edilebilir. </p> <p>Yani sayı çizgisi üzerindeki açık parantezler (-5) ve (0) arasındaki sayıları kapsar. </p>
Absolute Value Expression Evaluation
La expresión mostrada en la imagen es: - | -7 | Primero vamos a determinar el valor absoluto de -7. El valor absoluto de un número es la distancia que hay desde ese número hasta el 0 en la recta numérica, sin considerar la dirección. Por lo tanto, el valor absoluto de -7 es 7, ya que está a 7 unidades de distancia del 0. Ahora que tenemos | -7 | = 7, sustituimos este valor en la expresión inicial: - | -7 | = -7 Por lo tanto, el valor de la expresión es -7.
Understanding Inequalities Around -43
The image shows an inequality with two blank rectangles and a number in between, written as "< -43 <". To solve it, you need to insert mathematical symbols to make the statement true. Since -43 is a negative number, to make the inequality true, you'd want to indicate that the numbers decreasing from -43 are on the left and the numbers increasing from -43 are on the right. This means that you would use the "greater than" symbol, ">", on the left and the "less than" symbol, "<", on the right to show that the numbers on the left are less than -43 and the numbers on the right are greater than -43. To correctly fill in the blanks, the inequality should read: "< -43 >" This expresses that -43 is greater than the numbers to its left and less than the numbers to its right. However, this statement doesn't follow conventional notation because usually the "less than" symbol point towards the smaller number, making such a completed statement generally incorrect. A correct mathematical statement that reflects the ordering of numbers around -43 would be: "... < -43 < ..." This means any number to the left of -43 is less than it, and any number to the right is greater, which corresponds to the standard number line orientation.
Solving Inequalities with Absolute Value
To solve the inequality \( |t - 75| \leq 15 \), we need to split it into two separate inequalities, because the absolute value of a number is the distance from zero on the number line, and it can be expressed as the value being either greater than or equal to 0 or less than or equal to 0. The two cases for \( |t - 75| \leq 15 \) are: 1. \( t - 75 \leq 15 \) 2. \( -(t - 75) \leq 15 \) or equivalently \( t - 75 \geq -15 \) For the first case (1): \( t - 75 \leq 15 \) Adding 75 to both sides of the inequality gives us: \( t \leq 90 \) For the second case (2): \( t - 75 \geq -15 \) Adding 75 to both sides of this inequality gives us: \( t \geq 60 \) So, the solution to \( |t - 75| \leq 15 \) is the set of values of \( t \) that satisfy both conditions: \( 60 \leq t \leq 90 \) This means that the value of \( t \) is between 60 and 90, inclusive.
Solving Absolute Value Inequalities
The inequality \( |t - 75| \leq 15 \) represents the values of \( t \) that are at most 15 units away from 75 on the number line, either less than or equal to 75 or greater than or equal to 75, but not further away than 15 units in either direction. To solve this absolute value inequality, we'll break it into two separate inequalities: 1. \( t - 75 \leq 15 \) 2. \( - (t - 75) \leq 15 \) which is the same as \( t - 75 \geq -15 \) Now, we solve each inequality: 1. \( t \leq 15 + 75 \) \( t \leq 90 \) 2. \( t - 75 \geq -15 \) \( t \geq -15 + 75 \) \( t \geq 60 \) Combining these two inequalities, we get: \[ 60 \leq t \leq 90 \] Therefore, the solution to the absolute value inequality is all real numbers \( t \) between 60 and 90, inclusive.
Solving Absolute Value Inequalities
To solve the inequality \( |c| - 1 \geq 11 \), let's first isolate the absolute value expression on one side. \( |c| - 1 \geq 11 \) Add 1 to both sides to get: \( |c| \geq 12 \) This inequality means that the distance of \( c \) from 0 on the number line is greater than or equal to 12. Absolute value inequalities like this one split into two cases because the variable inside the absolute value can be either positive or negative. Case 1: If \( c \) is positive or zero, then we remove the absolute value: \( c \geq 12 \) Case 2: If \( c \) is negative, the absolute value makes it positive, so we have: \( -c \geq 12 \) Multiply both sides by -1, remembering to reverse the inequality sign since multiplying an inequality by a negative number reverses the direction of the inequality: \( c \leq -12 \) Now, we combine both cases into a compound inequality: \( c \leq -12 \) or \( c \geq 12 \) These inequalities represent all the possible values of \( c \) that make the original inequality true. In interval notation, this solution would be: \( (-\infty, -12] \cup [12, \infty) \)
Graphing an Inequality on a Number Line
The inequality given in the image is \( t \geq -3 \). To graph this inequality on a number line: 1. Locate the point -3 on the number line. 2. Since the inequality includes "greater than or equal to" (as indicated by the symbol \(\geq\)), you need a solid circle or dot at -3. This shows that -3 is part of the solution set. 3. Shade the number line to the right of -3, indicating all numbers greater than -3 are also included in the solution set. Now let's examine the provided options to see which graph corresponds to these instructions: A. This graph shows a number line with a solid dot at -3 and shading towards the left, which means values less than -3. This does not match the inequality. B. This graph shows a number line with an open circle at -3 (indicating that -3 is not included) and shading towards the right. This is not correct because the inequality specifies that -3 is included. C. This graph shows a number line with a solid dot at -3 and shading to the right, which seems to match the inequality \( t \geq -3 \). D. This graph depicts a number line with a solid dot at 3 and shading to the right; however, this does not correspond to the inequality provided. The correct answer is C, as it correctly represents the inequality \( t \geq -3 \).
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